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Chapter 9: Problem 38

If common sugars are heated too strongly, they char as they decompose into carbon and water vapor. For example, if sucrose (table sugar) is heated, the reaction is $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s) \rightarrow 12 \mathrm{C}(s)+11 \mathrm{H}_{2} \mathrm{O}(g) $$ What mass of carbon is produced if 1.19 g of sucrose decomposes completely?

Short Answer

Expert verified
To find the mass of carbon when 1.19 g of sucrose decomposes completely, we first calculate the moles of sucrose: \(0.00347 \, mol = \frac{1.19 \, g}{342.34 \, g/mol}\). Using stoichiometry, we find that the moles of carbon produced are \(0.04164 \, mol\). Converting this to mass using the molar mass of carbon, we get the mass of carbon produced as approximately \(0.50 \, g\).

Step by step solution

01

Write down the balanced chemical equation

We are given the balanced chemical equation as: \[ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s) \rightarrow 12\mathrm{C}(s) + 11 \mathrm{H}_{2} \mathrm{O}(g) \]
02

Calculate molar mass of sucrose and carbon

To determine the mass of carbon produced, we will first need to find out the molar mass (or molecular weight) of both sucrose and carbon. Molar mass of sucrose (\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)): - Carbon (C): \(12 \times 12.01 \, g/mol \approx 144.12 \, g/mol\) - Hydrogen (H): \(22 \times 1.01 \, g/mol \approx 22.22 \, g/mol\) - Oxygen (O): \(11 \times 16.00 \, g/mol \approx 176.00 \, g/mol\) Total molar mass of sucrose = \(144.12 + 22.22 + 176.00 \, g/mol \approx 342.34 \, g/mol\) Molar mass of carbon (C): \(12.01 \, g/mol\)
03

Determine the moles of sucrose

We are given the mass of sucrose to be 1.19 g. To find the moles of sucrose, we will use the following formula: \[moles = \frac{mass}{molar \, mass}\] So, moles of sucrose = \(\frac{1.19 \, g}{342.34 \, g/mol} \approx 0.00347 \, mol\)
04

Use stoichiometry to determine the moles of carbon produced

In the balanced chemical equation, we see that 1 mole of sucrose decomposes to produce 12 moles of carbon. Hence, we can calculate the moles of carbon produced using the moles of sucrose found in step 3. Moles of carbon = \(0.00347 \, mol \times 12 = 0.04164 \, mol\)
05

Calculate the mass of carbon produced

Now that we have the moles of carbon produced, we can calculate the mass of carbon using the molar mass of carbon. Mass of carbon = \(0.04164 \, mol \times 12.01 \, g/mol \approx 0.500 \, g\) So, if 1.19 g of sucrose decomposes completely, the mass of carbon produced is approximately 0.50 g.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction
A chemical reaction involves the conversion of one or more substances called reactants into one or more new substances known as products. The decomposition of sucrose when heated is an example of such a reaction, where it breaks down into carbon and water vapor. In educational contexts, understanding chemical reactions is fundamental as it helps in predicting the outcomes of reactions and establishing the relationship between reactants and products.

When you encounter a problem involving a chemical reaction, it is essential to comprehend the reactants' transformation into products and the way a reaction is represented through a chemical equation. In our example, the chemical process illustrates a type of reaction known as decomposition, where a complex molecule like sucrose breaks down into simpler elements and compounds under heat.
Molar Mass Calculation
The concept of molar mass is critical in stoichiometry as it allows for the conversion between grams and moles, which is key in measuring substances at the molecular level. Molar mass is calculated by adding up the atomic masses of all the atoms present in a molecule, typically expressed in grams per mole (g/mol).

To determine the molar mass of sucrose, or any compound, you must identify all the elements that make up the compound and their quantities. You then multiply the atomic mass of each element by the number of times it occurs in the molecule. Summing these values provides the molar mass of the compound, enabling further calculations in the problem-solving process. For instance, knowing the molar mass of sucrose helps us calculate how many moles of sucrose are present in a given mass, which is foundational for moving forward to find the mass of the products.
Balanced Chemical Equation
A balanced chemical equation is imperative for quantitatively predicting the substances formed during a chemical reaction. The principle of conservation of mass requires that the number of atoms of each element must be the same on both the reactant and product sides of a chemical equation.

In the given sucrose decomposition reaction, the equation showcased is already balanced, meaning that the atoms on both sides are equal in number. This balance allows us to use stoichiometry to relate the moles of reactants to the moles of products. Understanding the stoichiometric coefficients, which in this case indicates that 1 mole of sucrose yields 12 moles of carbon, is crucial when performing calculations. Knowing the balanced equation is the starting point for any stoichiometric problem and ensures the accuracy of your calculations.

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Most popular questions from this chapter

When yeast is added to a solution of glucose or fructose, the sugars are said to undergo fermentation, and ethyl alcohol is produced. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq}) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq})+2 \mathrm{CO}_{2}(g) $$ This is the reaction by which wines are produced from grape juice, Calculate the mass of ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), produced when \(5.25 \mathrm{~g}\) of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{0},\) undergoes this reaction.

Small quantities of ammonia gas can be generated in the laboratory by heating an ammonium salt with a strong base. For example, ammonium chloride reacts with sodium hydroxide according to the following balanced equation: $$ \mathrm{NH}_{4} \mathrm{Cl}(s)+\mathrm{NaOH}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ What mass of ammonia gas is produced if \(1.39 \mathrm{~g}\) of ammonium chloride reacts completely?

When elemental carbon is burned in the open atmosphere, with plenty of oxygen gas prescnt, the product is carbon dioxide. $$ \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) $$ However, when the amount of oxygen present during the burning of the carbon is restricted, carbon monoxide is more likely to result. $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}(g) $$ What mass of each product is expected when a \(5.00-\mathrm{g}\) sample of pure carbon is burned under cach of these conditions?

For cach of the following balanced reactions, calculate how many moles of each product would be produced by complete conversion of 0.50 mole of the reactant indicated in boldface. Indicate clearly the mole ratio used for the conversion. a. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) c. \(2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2}(g)\) d. \(\mathbf{C}_{3} \mathbf{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

Lead(II) carbonate, also called "white lead," was formerly used as a pigment in white paints. However, because of its toxicity, lead can no longer be used in paints intended for residential homes. Lead(II) carbonate is prepared industrially by reaction of aqueous lead(II) acetate with carbon dioxide gas. The unbalanced equation is $$ \mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{PbCO}_{3}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) $$ Suppose an aqueous solution containing 1.25 g of lead(II) acctate is treated with 5.95 g of carbon dioxide. Calculate the theoretical yield of lead carbonate.
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